Linear Inequalities/Inequations Question

mttal24 · 2012-05-06 22:15:29

My question,

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I understood that, all terms have to go on LHS, to solve in form:

1. But how to proceed ahead? Please explain me step-by-step.
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2. When you reach the end, you get two values(here, fractions), say c and d. Then how do you decide whether:

or

Similarly,

or

bobbym · 2012-05-06 22:48:20

Hi;

http://www.mathsisfun.com/algebra/inequ … lving.html

Here is some help:

x = 3 / 2 is a zero of the numerator and x =1 is undefined.

Bob · 2012-05-07 07:20:19

hi mttal24

Welcome to the forum!

If the question was

it would be easy to multiply by 3 and add 1 to both sides to isolate the 'x'.

The rules of algebra for equations would work fine for this inequality.

But with inequalities you always need to be on the alert for situations where you are multiplying (or dividing) by a negative.

Consider an easy example:

Add x to each side and take 5 from each side and this becomes

But look what happens if I try to do it by multiplying both sides by -1 and leave the < sign unchanged

This is incorrect! It happens because I have ignored the extra rule for inequalities

"If you multiply or divide by a negative, the inequality sign must be reversed."

So how does that effect your question?

You can make this

only if x-1 is not negative.

So let's assume x-1 > 0 and proceed

So, in this case, we want x > 1 AND x ≥ 1.5 Together these two are satisfied by x ≥ 1.5

But now consider the case where we are multiplying by a negative

ie x - 1 < 0 => x < 1

The two restrictions on x here may be replaced by the tougher condition x < 1

So there are two regions that make solution sets for the inequality

x < 1 and x ≥ 1.5

Bob

mttal24 · 2012-05-07 19:33:53

Thanks...
One thing more,
Can you show me a detailed procedure on how to solve inequations with the moduluses like | x-1 |

Bob · 2012-05-07 20:10:54

hi mttal24,

I find it helps to have a graph first. Let's say you are trying to solve

Start with the graph of y = x - 1

Now convert this into the graph of y = |x - 1|

Where y = x - 1 crosses the x axis there will be a change of direction as |x - 1| has no negative values.

You always end up with a V shaped graph. (see diagram below)

Then draw in the line y = 3 and you can see the solution range.

To do this analytically solve two separate equations

x - 1 = 3 gives the right end limit.

1 - x = 3 gives the left end limit.

So we end up with

Bob

mttal24 · 2012-05-07 21:13:56

^^ thanks. But how to do a little faster without a graph? I am seeing books which use zeroes of the linear equations given in numerator and denominator enclosed in the mods and then proceed.

Bob · 2012-05-07 22:08:18

Sketch of graph is usually sufficient. It doesn't have to be an accurate plot.

Post an example of a question and I'll say how I'd do it.

Bob

anonimnystefy · 2012-05-08 01:20:07

You can also see intervals in which the expession inside the modulus brackets changes sign.

mttal24 · 2012-05-08 16:50:51

@bob bundy
I have two for you:

1. |x-1|/x+2<1

2. |x-1|+|x-2| ≥4

Bob · 2012-05-08 18:50:00

hi mttal24

LATER EDIT: WHOOPS. I'VE SWITCHED A SIGN HERE AND ENDED UP DOING A SLIGHTLY DIFFERENT QUESTION.

(1) I'll consider three separate cases.

a) If x = 1. |x -1| = 0 so the left hand side is certainly < 1

b) If x > 1 |x -1| > 0 so we may ignore the || signs.

Between x = 1 and x = 2, x-2 is negative, so I need to split this into two cases.

b1) x > 1 and x < 2 (the inequality sign changes as I multiply by the negative)

This is true so there are solutions here.

b2) x > 2. the inequality sign stays as it is

This cannot happen so there are no solutions for x > 2

c) If x < 1 |x -1| < 0 so we can replace this expression with 1 - x and drop the || signs

furthermore if x < 1 then x - 2 < 0 so when I multiply by x - 2 I must reverse the inequality sign.

I now have two restrictions on x with x < 1 automatically causing the second to be true, so x < 1 is the range.

Taking a) b1) and c) together we get x < 2

The function is undefined at x = 2

Bob

ps. I've got to go out now for a while. I'll look at the other one when I get back.

Bob

mttal24 · 2012-05-08 20:23:08

^^Thanks.
In the book, the answer given is
soln. set of the inequation is (-infinity,-2) union (-1/2, +infinity).
But, you have given x<2 which would mean (-infinity, 2)

Last edited by mttal24 (2012-05-08 22:10:28)

Bob · 2012-05-08 23:41:08

hi mtall24

you have given x<2 which would mean (-infinity, 2)

Yes, these are the same. Brackets like this () mean the end points are not included. If they are included you could do [].

(2) |x-1|+|x-2| ≥4

The key points here are at x = 1 and x = 2.

So I'd consider 5 situations.

a) x = 1 You get 0 + 1 which is NOT ≥ 4

b) x = 2 You get 1 + 0 so again this is NOT part of the solution.

c) x < 1 Both parts of the || are negative so you can replace the || like this

1 - x + 2 - x ≥ 4
3 - 2x ≥ 4
-1 ≥ 2x
-1/2 ≥ x

d) 1 < x < 2 The first || is Ok; the second must be switched

x - 1 + 2 - x ≥ 4
1 ≥ 4 is NOT TRUE so no solutions here.

e) x > 2 Both expressions are positive so may be left unchanged.

x - 1 + x - 2 ≥ 4
2x - 3 ≥ 4
2x ≥ 7
x ≥ 3.5

To summarise x ≤ -1/2 AND x ≥ 3.5

Using brackets you could write this as

(- ∞, -0.5] AND [3.5, ∞)

Bob

anonimnystefy · 2012-05-09 00:28:17

Hi Bob

I think he is trying to say that the book got -2 and you got 2. Btw,the denominator of the expressiom is x+2.

Bob · 2012-05-09 00:33:26

hi Stefy,

I see what you mean.

I keep mis-reading questions. Glasses Ok; just the brain going screwy.

Hopefully the method is OK and, with a model answer, he can do the correct one himself.

Bob

anonimnystefy · 2012-05-09 00:40:47

bob bundy wrote:

just the brain going screwy.

The mouth on you! They should wash your tongue with a soap!

I think the method is okay. I don't,however,understand how they got one end to be at -1/2.

Stefy

Last edited by anonimnystefy (2012-05-09 00:41:05)

Bob · 2012-05-09 02:37:36

I've re-worked (1) with the plus sign and now I get

x < -2 and x > -1/2

Graph checks out as well.

And with post #11 tick.

Bob

mttal24 · 2012-05-09 17:08:07

Thanks. Finally I understood something...

mttal24 · 2012-05-09 18:04:13

One last thing please,
Can you explain how to solve
|x-2|/x-2 ≥ 0
My ans is:x ≥2
But book's ans is: x>2

Bob · 2012-05-09 19:35:04

hi mttal24

If you got that it looks like you are getting the hang of it. Well done!

The reason for x > 2 rather than x ≥ 2 is because the function is undefined at x = 2.

At this value, both the numerator and denominator are zero so it is not possible to fix a value.

So it is necessary to exclude x = 2 from the solution.

If you look at the graph, you can see that there is no clear value at x = 2

Bob

mttal24 · 2012-05-09 21:00:06

That means that the fx is undefined for denominator=0 because anything/0=undefined

Bob · 2012-05-09 21:13:59

Yes, that's it

If it's (non zero number) / zero, then this 'zooms' to infinity and that's not regarded as a 'value'.

If it's zero / zero then it could be anything because of the basic definition of division as the inverse of multiplication.

ie. If a x b = c , then a = c / b. this defines how division behaves.

But what if b and c are both zero?

Then we have

a x 0 = 0

There is no way you can say what 'a' is, so the full definition for division has to include " provided b is not zero "

Bob

Math Is Fun Forum

#1 2012-05-06 22:15:29

Linear Inequalities/Inequations Question

#2 2012-05-06 22:48:20

Re: Linear Inequalities/Inequations Question

#3 2012-05-07 07:20:19

Re: Linear Inequalities/Inequations Question

#4 2012-05-07 19:33:53

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#5 2012-05-07 20:10:54

Re: Linear Inequalities/Inequations Question

#6 2012-05-07 21:13:56

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#7 2012-05-07 22:08:18

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#8 2012-05-08 01:20:07

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#9 2012-05-08 16:50:51

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#10 2012-05-08 18:50:00

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#11 2012-05-08 20:23:08

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#12 2012-05-08 23:41:08

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#15 2012-05-09 00:40:47

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#17 2012-05-09 17:08:07

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